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An engine has an efficiency of 0.25 when...

An engine has an efficiency of 0.25 when temperature of sink is reduced by `58^(@)C` , If its efficiency is doubled, then the temperature of the source is

A

`150 ^(@)C`

B

`222 ^(@)C`

C

`242 ^(@)C`

D

`232 ^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
D
Here `eta_(1)=1-(T_(2))/(T_(1))`
or 0.25 =1 -`(T_(2))/(T_(1)) rArr (1)/(4) =1-(T_(2))/(T_(1))`
`(T_(2))/(t_(1))=1-(1)/(4)=(3)/(4)`
According to question,
`eta_(2)=2eta_(1)` and `T_(2)=T_(2)-58^(@)C`
`therefore 2xx(1)/(4)=1(T_(2)-58^(@)C)/(T_(1)) rArr (1-(1))/(2)=(T_(2)-58^(@)C)/(T_(1))`
`(1)/(2)=(T_(2))/(T_(1))-(58^(@))/(t_(1))rArr(3)/(4)-(1)/(2)=(58)/(T_(1)) rArr T_(1)=232^(@)C`
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